10E, 10F, 10H - Trigonometric equations using identities

How can an equation involving both sine and cosine turn into an ordinary quadratic? You will use the Pythagorean identity (10E) and tangent quotient identity (10F) to solve the interval equations of 10H, recovering every permitted angle. You will also keep zero-factor solutions, both signs of a square root and restrictions from the original expression.

Quadratics in a trig function

The expression $2\sin^2x+\sin x-1$ has the same algebraic structure as $2s^2+s-1$ . Replacing the repeated ratio by a temporary variable makes that structure visible. Solving for the ratio is only the first stage: the question asks for angles.

When sine and cosine both occur, $\sin^2x+\cos^2x=1$ can remove one squared ratio. Choose the form that leaves one function, such as $\cos^2x=1-\sin^2x$ when an unsquared sine term is already present.

Problem-solving

Solve the algebraic equation in $s=\sin x$ , $c=\cos x$ or $t=\tan x$ . Reject sine or cosine values outside $[-1,1]$ , then solve each remaining trig equation on the given interval. Tangent has no corresponding bound on its real values.

Worked example 1: using an identity before factorising

Question. Solve $2\cos^2x-\sin x-1=0$ for $0^\circ\le x<360^\circ$ .

Plan. Replace $\cos^2x$ by $1-\sin^2x$ , then form a quadratic in $s=\sin x$ .

\begin{aligned} 2(1-\sin^2x)-\sin x-1&=0\\ 2\sin^2x+\sin x-1&=0\\ (2s-1)(s+1)&=0 \end{aligned}

Thus $\sin x=1/2$ or $\sin x=-1$ . Both ratio values are possible.

\sin x=\frac12\Rightarrow x=30^\circ,150^\circ; \qquad\sin x=-1\Rightarrow x=270^\circ.

Answer. $x=30^\circ,150^\circ,270^\circ$ .

Check. At the first two angles the original left side is $2(3/4)-1/2-1=0$ . At $270^\circ$ it is $0-(-1)-1=0$ . All angles are in the interval.

The next equation is already in one ratio. Its algebra has two roots, but only one can be a cosine value.

Worked example 2: rejecting an impossible ratio

Question. Solve $2\cos^2x-5\cos x+2=0$ for $-\pi\le x\le\pi$ .

Plan. Put $c=\cos x$ , factorise, then use radian cosine solutions.

2c^2-5c+2=(2c-1)(c-2)=0 \quad\Rightarrow\quad c=\frac12\text{ or }2.

Reject $c=2$ , since a real cosine cannot exceed $1$ . For $\cos x=1/2$ , the solutions in the interval are $x=\pm\pi/3$ .

Answer. $x=-\pi/3,\pi/3$ .

Check. At either angle the original expression is $2(1/4)-5(1/2)+2=0$ . The endpoints have cosine $-1$ , which gives $9$ , not zero.

Complete both stages in the next equation.

Factorising without losing solutions

Dividing by $\sin x$ assumes that $\sin x\ne0$ . If the original equation permits $\sin x=0$ , that division removes genuine solutions. Move everything to one side and factor instead: a product is zero when at least one factor is zero.

Worked example 3: keeping the zero factor

Question. Solve $\sin x\cos x=\tfrac12\sin x$ for $0^\circ\le x\le360^\circ$ .

Plan. Retain the common sine factor rather than cancelling it.

\sin x\cos x-\frac12\sin x=0 \quad\Rightarrow\quad\sin x\left(\cos x-\frac12\right)=0.

So $\sin x=0$ or $\cos x=1/2$ .

\sin x=0\Rightarrow x=0^\circ,180^\circ,360^\circ; \qquad\cos x=\frac12\Rightarrow x=60^\circ,300^\circ.

Answer. $x=0^\circ,60^\circ,180^\circ,300^\circ,360^\circ$ .

Check. The three sine-zero angles make both sides zero. At the other two angles multiplying $\sin x$ by $\cos x=1/2$ gives exactly the right side. Both endpoints are included.

When tangent appears, there is an extra domain decision before multiplying by a denominator.

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10E, 10F, 10H - Trigonometric equations using identities

Quadratics in a trig function

Factorising without losing solutions

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